Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &s, vector<T> vec) {
  for (auto &v : vec) {
    s << v << " ";
  }
  return s;
}
template <typename T>
void print(T v) {
  cout << v << "\n";
}
template <typename T>
void prints(T v) {
  cout << v << "\n";
}
template <typename T, typename... Args>
void print(T first, Args... args) {
  cout << first << " ";
  print(args...);
}
template <typename T, typename... Args>
void prints(T first, Args... args) {
  cout << first << " | ";
  prints(args...);
}
int rowCheck(const vector<vector<int>> &mat, const int &r) {
  int minVal = 1000;
  for (int c = 0; c < mat[r].size(); ++c) {
    minVal = min(minVal, mat[r][c]);
  }
  return minVal;
}
void rowModify(vector<vector<int>> &mat, const int &r, const int &minVal) {
  for (int c = 0; c < mat[r].size(); ++c) {
    mat[r][c] -= minVal;
  }
}
int colCheck(const vector<vector<int>> &mat, const int &c) {
  int minVal = 1000;
  for (int r = 0; r < mat.size(); ++r) {
    minVal = min(minVal, mat[r][c]);
  }
  return minVal;
}
void colModify(vector<vector<int>> &mat, const int &c, const int &minVal) {
  for (int r = 0; r < mat.size(); ++r) {
    mat[r][c] -= minVal;
  }
}
void rowFunction(vector<vector<int>> &mat, stringstream &ss, int &count) {
  for (int r = 0; r < mat.size(); ++r) {
    int minVal = rowCheck(mat, r);
    if (minVal) {
      rowModify(mat, r, minVal);
      for (int t = 0; t < minVal; ++t) {
        ss << "row " << r + 1 << "\n";
      }
      count += minVal;
    }
  }
}
void colFunction(vector<vector<int>> &mat, stringstream &ss, int &count) {
  for (int c = 0; c < mat[0].size(); ++c) {
    int minVal = colCheck(mat, c);
    if (minVal) {
      colModify(mat, c, minVal);
      for (int t = 0; t < minVal; ++t) {
        ss << "col " << c + 1 << "\n";
      }
      count += minVal;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> mat(n, vector<int>(m, 0));
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) cin >> mat[i][j];
  stringstream ss;
  int count = 0;
  if (n > m) {
    colFunction(mat, ss, count);
    rowFunction(mat, ss, count);
  } else {
    rowFunction(mat, ss, count);
    colFunction(mat, ss, count);
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (mat[i][j]) {
        cout << "-1";
        return 0;
      }
    }
  cout << count << "\n";
  cout << ss.str();
}